One day, two friends named Al and Bob found a shiny quarter on the ground. To decide who would get it, Al would flip the coin. If it came up heads, then Al would keep it, but if it was tails, Bob would have to flip for it. If Bob got heads, Bob would keep the coin, but if he got tails, then Al would flip it and the entire process would repeat. Assuming that Al goes first and that the coin has an equal chance of landing on either heads or tail, what is the probability that Al would keep the coin?

Hint

Answer

2/3

If Al flipped heads, then he would win, and the probability of that happening would be 1/2. However, he could also win if he flipped tails, Bob flipped tails, and Al flipped heads afterwards. The probability of this happening would be:

(1/2)(1/2)(1/2)=1/8

If this process were to continue indefinitely, then the probability of Al winning would be:

(1/2)+(1/8)+(1/32)+...+(1/2)(1/4)^n

in which n equals an integer one greater than the end preceding it and one less than one integer after it. Using the formula to obtain a solution for the summation of a geometric sequence, it is possible to deduce that the summation equals: